Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 30

MATH152 April 2015

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Question A 30
Consider the probability transition matrix $P={\begin{bmatrix}1/2&0&0&0\\0&1&0&1/3\\1/2&0&1&0\\0&0&0&2/3\end{bmatrix}}$ and initial probability $\mathbf {x} _{0}={\begin{bmatrix}1/11\\3/11\\3/11\\4/11\end{bmatrix}}$ What is $\lim _{n\to \infty }P^{n}\mathbf {x} _{0}$? Hint: this can be done without extensive calculations.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Think about where each of the individual states is likely to end up after several iterations.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let us examine what happens to the individual states $${\mathbf {e_{1}}}={\begin{bmatrix}1\\0\\0\\0\end{bmatrix}},\quad {\mathbf {e_{2}}}={\begin{bmatrix}0\\1\\0\\0\end{bmatrix}},\quad {\mathbf {e_{3}}}={\begin{bmatrix}0\\0\\1\\0\end{bmatrix}},\quad {\mathbf {e_{4}}}={\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}.$$ Examining ${\textstyle {\mathbf {e_{2}}}}$, we can see the 2nd column of ${\textstyle P}$ is $${\begin{bmatrix}0\\1\\0\\0\end{bmatrix}}$$ so the second state is stationary (${\textstyle P^{n}{\mathbf {e_{2}}}={\mathbf {e_{2}}}}$ for all ${\textstyle n}$). Likewise, ${\textstyle P^{n}{\mathbf {e_{3}}}={\mathbf {e_{3}}}}$. Examining ${\textstyle {\mathbf {e_{1}}}}$, we can see that every iteration this state has probability ${\textstyle 1/2}$ of becoming third state and ${\textstyle 1/2}$ of remaining the same (by the first column of ${\textstyle P}$). Since ${\textstyle {\mathbf {e_{3}}}}$ is stationary, this means that $$P^{n}{\mathbf {e_{1}}}={\frac {1}{2^{n}}}{\mathbf {e_{1}}}+\left(1-{\frac {1}{2^{n}}}\right){\mathbf {e_{3}}}$$ , so as ${\textstyle n}$ goes to infinity, the probability of ${\textstyle {\mathbf {e_{1}}}}$ becoming the third state goes to one. Similarly, as the number of iterations increases, the probability of ${\textstyle {\mathbf {e_{4}}}}$ becoming the second state goes to one. Putting this all together, we have that the probability of ending up in the second state as ${\textstyle n\rightarrow \infty }$ is the sum of the probabilities of starting in the second and fourth states, i.e., ${\textstyle {\frac {7}{11}}}$, while by the same reasoning the probability of ending up in the third state is ${\textstyle {\frac {4}{11}}}$ (the remaining states have probability zero). This means our answer is $$\lim _{n\rightarrow \infty }P^{n}{\mathbf {x_{0}}}=\color {blue}{\begin{bmatrix}0\\7/11\\4/11\\0\end{bmatrix}}.$$